Đáp án:
Giải thích các bước giải:
Ta có:
$sin^4a+cos^4a-1$
$=sin^4a+2.sin^2acos^2a+cos^4a-1-2sin^2acos^2a$
$=(sin^2a+cos^2a)^2-1-2sin^2acos^2a$
$=1^2-1-2sin^2acos^2a$
$=-2sin^2acos^2a$
Lại có:
$sin^6a+cos^6a-1$
$(sin^2a)^3+(cos^2a)^3-1$
$=(sin^2a+cos^2a)(sin^4a-sin^2acos^2a+sin^2a)-1$
$=1(sin^4a+2sin^2acos^2a+sin^2a-3sin^2acos^2a)-1$
$=(sin^2a+cos^2a)^2-1-3sin^2acos^2a$
$=1^2-1-3sin^2acos^2a$
$=3sin^2acos^2a$
Ta có:
`\frac{sin^4a+cos^4a-1}{sin^6a+cos^6a-1}`
`=\frac{-2sin^2acos^2a}{-3sin^2acos^2a}`
`=\frac{2}{3}` (đpcm)
